The opponents of soccer team A are of two types: either they are a class 1 or a class 2 team. The number of goals team A scores against a class i opponent is a Poisson random variable with mean λ , where λ = 2, λ = 3. This weekend the team has two games against teams they are not very familiar with. Assuming that the first team they play is a class 1 team with probability 0.6 and the second is, independently of the class of the first team, a class 1 team with probability 0.3, determine
(a) the expected number of goals team A will score this weekend.
(b) the probability that team A will score a total of five goals.

I need help with part b of this question. I have part a covered. Please explain in detail on how to obtain the answer for part b.

a) The expected number of goals team A will score is 5.1

b) The probability that team A will score a total of 5 goals is 0.1147

Step-by-step explanation:

Let X be the amount of goals scored by team A in both matches. Let X1 and X2 be the total amount of goals team A scores in match 1 and 2 respectively, then X = X1+X2, and also

a)

E(X) = E(X1+X2) = E(X1)+E(X2) = 0.6*2+0.4*3 + 0.3*2+0.7*3 = 5.1

b) In order for X to be equal to 5 we have 5 possibilities

- X1 is 0 and X2 is 5

- X1 is 1 and X2 is 4

- X1 is 2 and X2 is 3

- X1 is 3 and X2 is 2

- X1 is 4 and X2 is 1

- X1 is 5 and X2 is 0

Let T1 be a poisson distribution with mean λ = 2, then

Lets do the same with a Poisson distribution T2 with mean λ = 3

Now, we are ready to compute the probability that X is equal to 5.

We can conclude that

P(X = 5) = 0.00823+0.03214+0.05317+0.0471+0.0225+0.0047 = 0.1147

The probability that team A will score a total of 5 goals is 0.1147

the 8 in 8,420 is in the thousands place so it's value is 8 thousand.

the 8 in 6,870 is in the hundreds place so it's value is 8 hundred.

divide 8000 by 800 to find the answer:

8000 / 800 = 10

the 8 in 8,420 is 10 times greater than the 8 in 6,870